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• Today, 17:34
Yes, correct.
2 replies | 27 view(s)
• Today, 10:34
Yes, this is correct.
2 replies | 35 view(s)
• June 22nd, 2017, 11:55
Here is this week's POTW: ----- Evaluate \sum_{i=1}^{1995}\dfrac{1}{f(i)}, given that f(k) be the integer closest to \sqrt{k}. ----- ...
0 replies | 42 view(s)
• June 22nd, 2017, 11:49
No one answered last week's problem.(Sadface) You can find the suggested solution below: The claim that the given system of equations is...
1 replies | 90 view(s)
• June 21st, 2017, 15:55
What do you mean with feedback loop?
9 replies | 93 view(s)
• June 21st, 2017, 15:35
So does this mean that if we get from $x_i$ to some gate that represents one specific operation $\star$ , will we then apply the operation $x_i... 9 replies | 93 view(s) • June 21st, 2017, 15:12 So if we have$x_1=1$and$x_2=0$do we have to take the first possible way for$x_2$, then we have the expression$x_1$AND$x_2$and get$1 \wedge...
9 replies | 93 view(s)
• June 21st, 2017, 14:54
But, why is it like that? Since at the circuit there aren't said any inputs, we can go anywhere with input either 0 or 1 ?(Thinking)
9 replies | 93 view(s)
• June 21st, 2017, 14:41
Hello!!! (Wave) I read the following about parity functions: Since I haven't heard of combinatorial circuits before, I have searched for...
9 replies | 93 view(s)
• June 16th, 2017, 16:28
Hi, and welcome to the forum. Suppose that $(x,y)\in A\times(B\triangle C)$. Then $x\in A$ and $y\in B\triangle C$. The latter means that $y\in B$...
1 replies | 79 view(s)
• June 15th, 2017, 11:55
Here is this week's POTW: ----- Determine $a^2+b^2+c^2+d^2$ if ...
1 replies | 90 view(s)
• June 15th, 2017, 11:43
Congratulations to kaliprasad for his correct solution, and you can find the suggested solution as follows::) Let the perpendicular distance from...
1 replies | 94 view(s)
• June 15th, 2017, 07:11
On this forum you are supposed to show an attempted solution or at least describe the difficulty. Surely you can plot the first graph: there is...
13 replies | 190 view(s)
• June 14th, 2017, 22:13
In the US curriculum, pre-calculus is a little bit vague in its goals but from my experience there are some key ideas that are covered that are...
4 replies | 88 view(s)
• June 13th, 2017, 04:40
Yes. It's impossible to express this equation in the form y = mx + b. This form represents a function y(x), and the line in question is not the...
6 replies | 110 view(s)
• June 13th, 2017, 04:33
Yes, you are correct. You may use the following equation of the line passing through (x1, y1) and (x2, y2). \ if y2 - y1 = 0, then the equation...
2 replies | 48 view(s)
• June 12th, 2017, 18:39
That sounds frustrating, karush. I am sorry that this has been a recurring problem and thank you for letting us know. From everything you've told...
16 replies | 285 view(s)
• June 12th, 2017, 13:56
In post #10 I wrote the following. Going back to the original equation x^2 + 2Ax + y^2 + 2By = C with x-intercepts a, b and y-intercepts c, d, you...
16 replies | 225 view(s)
• June 12th, 2017, 12:23
I am not sure what you mean (LaTeX formulas are also typed). If my post is displayed incorrectly, the best thing would be to post a screenshot to...
16 replies | 225 view(s)
• June 12th, 2017, 12:11
Going back to the original equation $x^2 + 2Ax + y^2 + 2By = C$ with $x$-intercepts $a$, $b$ and $y$-intercepts $c$, $d$, you can express $ab$ and...
16 replies | 225 view(s)
• June 12th, 2017, 12:02
That's why I provided a link. In my opinion, this information is useful, at least to verify the answers, i.e., formulas for sum and product of roots...
16 replies | 225 view(s)
• June 12th, 2017, 10:53
Mark is offering you a proof of Vieta's formulas (see example for application to quadratic polynomials).
16 replies | 225 view(s)
• June 10th, 2017, 10:28
Evgeny.Makarov replied to a thread Jokes in Chat Room
160 replies | 19457 view(s)
• June 10th, 2017, 09:38
So we are talking about the claim \ for all $x$. I suggested checking it for $x=-1$. You determined that $|x|>0$ is true because...
10 replies | 141 view(s)
• June 9th, 2017, 14:56
No, I prefer you check first whether you claim is true for $x=-1$. We are talking about your claim: $|x|>0\iff x\text{ lies to the right of 0}$...
10 replies | 141 view(s)
• June 9th, 2017, 13:12
You are claiming that $|x| > 0$ iff the point $x$ lies to the right 0 for all $x$. I suggest you check this claim for $x=-1$. If a person says,...
10 replies | 141 view(s)
• June 9th, 2017, 04:43
Yes. What about $x=-1$?
10 replies | 141 view(s)
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